Jee Main 2024 29 01 2024 Shift 1 - Question1
Question 1
A body of man $100 kg$ travelled $10 m$ before coming to rest. If $\mu=0.4$, work done against friction is (motion is happening on horizontal surface, take $g=10 m / s^{2}$ )
(1) $4500 J$
(2) $5000 J$
(3) $4200 J$
(4) $4000 J$
Show Answer
Answer: (4)
Solution:
$\frac{v^{2}}{2 a}=s$
$(a=\mu g)$
$v^{2}=2 \mu g s$
$v^{2}=2 \times 0.4 \times 10 \times 10$
$v^{2}=80$
$w _f=\Delta k$
$=-\frac{1}{2} \times 100 \times 80$
$w _f=-4000$
BETA
